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Chapter 32: Problem 17

A space probe \(2.0 \times 10^{10} \mathrm{~m}\) from a star measures the total intensity of electromagnetic radiation from the star to be \(5.0 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2}\). If the star radiates uniformly in all directions, what is its total average power output?

Short Answer

Expert verified
To calculate the total average power output of the star, you first calculate the spherical surface area at the location of the probe. Using this area and the given intensity, you find the total power output by multiplying the intensity and the surface area.

Step by step solution

01

Understand the problem and what is given

The intensity \(I\) of the electromagnetic radiation is \(5.0 \times 10^{3} \mathrm{~W/m^{2}}\) where the distance \(r\) between the star and the probe is \(2.0 \times 10^{10} \mathrm{~m}\). The problem involves calculating the total power \(P\) output of the star.
02

Apply the equation for the surface area of a sphere

The surface area \(A\) of a sphere can be described as \(A = 4\pi r^2\). Let's put the value of \(r\) into the equation: \(A = 4\pi (2.0 \times 10^{10} \mathrm{~m})^2\).
03

Calculate the total power output

Power \(P\) can be described as the product of intensity \(I\) and the area \(A\) over which it is spread. Thus the total power output of the star will be \(P = IA\). Substituting \(I\) and \(A\) into the equation will give us the total power output.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Output
Power output is a critical concept in physics, especially when discussing electromagnetic radiation from stars and other celestial bodies. When we mention power output, we mean the total amount of energy a star emits per unit time. This is usually measured in watts (W). It represents how much energy is distributed into the surrounding space and affects various factors, such as the light intensity reaching planets and probes.

Consider a star radiating energy uniformly in all directions. The power output can help us understand the vastness of its energy emission. The more powerful a star, the higher its ability to illuminate and heat surrounding areas. This understanding is key when studying how stars influence their nearby planets and their ability to support life.

In our context, power output is derived from knowing the intensity (energy per unit area) and applying it to the geometry of a sphere, which will be our next concept. This transformation helps simplify many astronomical calculations.
Intensity Calculation
Intensity is a measure of how much power is received per unit area, measured in \( ext{W/m}^{2} \). When dealing with electromagnetic radiation, like light from a star, intensity tells us how dense or concentrated this energy is over a given surface area. This density of energy is crucial in understanding how bright or warm an object receiving this radiation will feel.

Calculating intensity involves knowing how much power is available and over what area it spreads. As stars emit energy in all directions, understanding this spread becomes essential. In our problem, we already know the measured intensity at a known distance from the star. This information allows us to calculate how much total power the star is emitting.

By working backwards from the measured intensity at a specific location (like our space probe position), we can determine the total power output of the star, which turns out to be a fundamental aspect of astronomy.
Surface Area of a Sphere
The surface area of a sphere plays a vital role when dealing with celestial objects radiating energy evenly in all directions, such as stars. Mathematically, the surface area \( A \) is given by the formula \( A = 4 \pi r^{2} \), where \( r \) is the radius of the sphere.

In this context, the sphere represents an imaginary shell at a fixed distance from the star, specifically the distance to the space probe. This formula becomes handy as it helps calculate how the star's emitted energy is distributed over space.

When you know this surface area, you can use it alongside intensity to find the total power output, as outlined in the exercise. By multiplying intensity by the surface area, you can derive the total energy the star emits, giving a clear insight into its energy properties. Understanding this calculation provides a stronger grasp of how energy disperses in space, impacting planets and other objects around the star.

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Most popular questions from this chapter

When electromagnetic radiation strikes perpendicular to a flat surface, a totally absorbing surface feels radiation pressure \(I_{0} / c\) where \(I_{0}\) is the intensity of incident electromagnetic radiation. A totally reflecting surface feels twice that pressure. More generally, a surface absorbs a proportion \(e\) of the incident radiation and reflects a complementary proportion, \(1-e,\) where \(e\) is the emissivity of the surface, as introduced in Chapter \(17 .\) Note that \(0 \leq e \leq 1 .\) (a) Determine the radiation pressure \(p_{\text {rad }}\) in terms of \(I_{0}\) and \(e .\) (b) Consider cosmic dust particles in outer space at a distance of \(1.5 \times 10^{11} \mathrm{~m}\) from the sun, where \(I_{\text {sun }}=1.4 \mathrm{~kW} / \mathrm{m}^{2}\). We can model these particles as tiny disks with \(e=0.61,\) diameter \(8.0 \mu \mathrm{m}\) and mass \(1.0 \times 10^{-10} \mathrm{grams},\) all oriented perpendicular to the sun's rays. What is the force on one of these particles that is exerted by the radiation from the sun? (c) What is the ratio of this force to the attractive force of gravity exerted by the sun on the particle?

An air-filled cavity for producing electromagnetic standing waves has two parallel, highly conducting walls separated by a distance \(L\). One standing- wave pattern in the cavity produces nodal planes of the electric field with a spacing of \(1.50 \mathrm{~cm}\). The next-higher-frequency standing wave in the cavity produces nodal planes with a spacing of \(1.25 \mathrm{~cm} .\) What is the distance \(L\) between the walls of the cavity?

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho\). (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M\) ) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r\). The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about \(3000 \mathrm{~kg} / \mathrm{m}^{3} .\) Find the particle radius \(R\) such that the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is \(3.9 \times 10^{26} \mathrm{~W}\). Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge \(q\) and acceleration \(a\) is given by $$ \frac{d E}{d t}=\frac{q^{2} a^{2}}{6 \pi \epsilon_{0} c^{3}} $$ where \(c\) is the speed of light. (a) Verify that this equation is dimensionally correct. (b) If a proton with a kinetic energy of \(6.0 \mathrm{MeV}\) is traveling in a particle accelerator in a circular orbit of radius \(0.750 \mathrm{~m},\) what fraction of its energy does it radiate per second? (c) Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?

We can reasonably model a \(75 \mathrm{~W}\) incandescent light bulb as a sphere \(6.0 \mathrm{~cm}\) in diameter. Typically, only about \(5 \%\) of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in \(\left.\mathrm{W} / \mathrm{m}^{2}\right)\) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?
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